Carbon monoxide 4 with a volume of 4.48 liters was passed through lime water, the precipitate formed

Carbon monoxide 4 with a volume of 4.48 liters was passed through lime water, the precipitate formed was separated and calcined, calculate the mass of the solid residue obtained after calcination.

Given:
V (CO2) = 4.48 l

To find:
m (remainder) -?

Decision:
1) Write the reaction equations:
Ca (OH) 2 + CO2 => CaCO3 + H2O;
CaCO3 => CaO + CO2;
2) Calculate the amount of substance CO2:
n (CO2) = V (CO2) / Vm = 4.48 / 22.4 = 0.2 mol;
3) Determine the amount of CaCO3 substance:
n (CaCO3) = n (CO2) = 0.2 mol;
4) Determine the amount of CaO substance:
n (CaO) = n (CaCO3) = 0.2 mol;
5) Calculate the molar mass of CaO:
M (CaO) = Mr (CaO) = Ar (Ca) * N (Ca) + Ar (O) * N (O) = 40 * 1 + 16 * 1 = 56 g / mol;
6) Calculate the mass of CaO:
m (CaO) = n (CaO) * M (CaO) = 0.2 * 56 = 11.2 g.

Answer: The mass of CaO is 11.2 g.